# Calculus 3 way simultaneous 'Find value of K such that there is no unique solutions'

#### brendanvondoom

Help, I'm lost on how to answer this;
Find value of K such that there is no unique solutions for the following system of equations.
3x + y - z = 1 (1)
x -y +kz =7 (2)
2x+ky-z=-1 (3)

DO not use that method/set up bellow. I want to solve simultaneously manipulating eqns...
3...1..-1
1..-1...k
2...k..-1

Must use simultaneous to solve for K... so far I got

(2)x3 -(1).............. -4y+3kz+z =20 (4)

(2)x2 - (3) -2y-ky+2kz-z=15 (5)

(5)x2 - (4) -2ky+kz-3z=10
now what? Hope My steps so far are right...

Note: I cannot use this method:
|3 1 -1|3 1|
|1 -1 k|1 -1| =
|2 k -1|2 k|

Cheers.
I've tried my best so far but I just can't seem to get it. Thanks

#### Prove It

MHF Helper
Re: Calculus 3 way simultaneous 'Find value of K such that there is no unique solutio

Help, I'm lost on how to answer this;
Find value of K such that there is no unique solutions for the following system of equations.
3x + y - z = 1 (1)
x -y +kz =7 (2)
2x+ky-z=-1 (3)

DO not use that method/set up bellow. I want to solve simultaneously manipulating eqns...
3...1..-1
1..-1...k
2...k..-1

Must use simultaneous to solve for K... so far I got

(2)x3 -(1).............. -4y+3kz+z =20 (4)

(2)x2 - (3) -2y-ky+2kz-z=15 (5)

(5)x2 - (4) -2ky+kz-3z=10
now what? Hope My steps so far are right...

Note: I cannot use this method:
|3 1 -1|3 1|
|1 -1 k|1 -1| =
|2 k -1|2 k|

Cheers.
I've tried my best so far but I just can't seem to get it. Thanks
In order for a system of equations to have a unique solution, it needs to have as many unknowns as equations (which it does) and the determinant of the coefficient matrix can not be 0. So if you are told that you can't have unique solutions, then the determinant IS 0. So evaluate the determinant, set it equal to 0, then solve for k.

#### brendanvondoom

Re: Calculus 3 way simultaneous 'Find value of K such that there is no unique solutio

Hi thanks. But how about would I get from my working to get K? My teacher has not taught me this 'evaluation of the determinent' or something as our standard, to get requires you to solve it like using the 3 equations... as shown in my working...

#### Soroban

MHF Hall of Honor
Re: Calculus 3 way simultaneous 'Find value of K such that there is no unique solutio

Hello, brendanvondoom!

Find value of $$\displaystyle k$$ such that there is no unique solutions for this system of equations.

. . $$\displaystyle \begin{array}{ccccc}3x + y - z &=& 1 & [1] \\ x -y +kz &=&7 & [2] \\ 2x+ky-z&=& \text{-}1 & [3] \end{array}$$

Must use simultaneous to solve for $$\displaystyle k$$.

So far I got:

. . $$\displaystyle \begin{array}{cccccc}3\!\cdot\![2] -[1]: & \text{-}4y+3kz+z &=&20 & [4] \\2\!\cdot\![2] - [3]: & \text{-}2y-ky+2kz\:{\color{red}+}\:z&=& 15 & [5] \end{array}$$

Hope my steps so far are right. . Almost!

Answers are: .$$\displaystyle k=1,\;\text{-}\tfrac{2}{3}$$

We have: .$$\displaystyle \begin{array}{cccccc}\text{-}4y &+& (3k+1)z &=& 20 & [4] \\ \text{-}(k+2)y &+& (2k+1)z &=& 15 & [5] \end{array}$$

$$\displaystyle \begin{array}{ccccccccc}(k+2)\!\cdot\![4]: & \text{-}4(k+2)y &+& (3k+1)(k+2)z &=& 20(k+2) \\ \text{-}4\!\cdot\![5]: & 4(k+2)y &-& 4(2k+1)z &=& \text{-}60 \end{array}$$

Add: .$$\displaystyle \big[(3x+1)(k+2) - 4(2k+1)\big]z \;=\;20(k+2) - 60$$

n . . . . . . . . . . . . . . $$\displaystyle (3k^2-k-2)z \;=\;20k+40 - 60$$

. . . . . . . . . . . . . . $$\displaystyle (k-1)(3k+2)z \;=\;20k - 20$$

. . . . . . . . . . . . . . . . . . . . . . . . $$\displaystyle z \;=\;\frac{20(k-1)}{(k-1)(3x+2)}$$

There is no unique solution if the denominator equals zero.

. . Therefore: .$$\displaystyle k = 1\,\text{ or }\, k = \text{-}\tfrac{2}{3}$$

1 person

#### brendanvondoom

Re: Calculus 3 way simultaneous 'Find value of K such that there is no unique solutio

Hello Soroban,
THANK YOU man so much Bro. Your working is really clear and so helpful man!.
Again Thank you for your time, much appreciated!